# What rules do I break transforming free index into dummy index in Einstein summation notation?

I got this equation for matrix element, no summation notation, just free indices: $$\varepsilon_{ij} = \dfrac{u_i}{k_j}+\dfrac{u_j}{k_i}$$

I want to sum all the elements of this matrix multiplied in some manner for which, I belive, I can use summation notation. From now on I want to use i,j as dummy indices: $$\varepsilon_{ij} = \dfrac{u_i}{k_j}+\dfrac{u_j}{k_i} \quad \mid \cdot k_ik_j \\ k_ik_j \varepsilon_{ij} = \dfrac{k_i k_ju_i}{ k_j}+\dfrac{k_ik_ju_j}{ k_i} = k_iu_i + k_ju_j$$

I guess it is legitimate.

Why can't I then write $$k_i u_i$$ as a vectors dot product: $$k_ik_j\varepsilon_{ij} = k_iu_i + k_ju_j = \vec k \cdot \vec u+ \vec k \cdot \vec u = 2\vec k \cdot \vec u$$ and get back to my initial matrix?

$$k_ik_j \varepsilon_{ij} = 2\vec k \cdot \vec u \quad \mid : (k_ik_j) \\ \varepsilon_{ij} = \dfrac{2\vec k \cdot \vec u }{k_ik_j }$$ This is obviously not true, as:

$$\varepsilon_{xz} = \dfrac{2\vec k \cdot \vec u }{k_xk_z } = 2\dfrac{k_xu_x+k_yu_y+k_zu_z}{k_xk_z} \neq \dfrac{u_x}{k_z}+ \dfrac{u_z}{k_x}$$

• Your last step is the problem- you can‘t divide by $k_ik_j$, because there is a sum over all $i$ and $j$ on the left side. Sep 1 '20 at 15:53
• Thank you! Please write an answer and I will accept it. Sep 1 '20 at 16:38

The short answer: Dividing the equation $$k_ik_j\varepsilon_{ij}=2\vec k\cdot\vec u$$ by $$k_ik_j$$ is not allowed, because you free the summation indices $$i$$ and $$j$$.
A little longer answer: The matrix definition $$\varepsilon_{ij} = \dfrac{u_i}{k_j}+\dfrac{u_j}{k_i}$$ consists of $$d^2$$ equations, when $$d$$ is the dimension. After summing up you are at the single equation $$k_ik_j\varepsilon_{ij}=2\vec k\cdot\vec u$$. There is no way to conclude on $$d^2$$ equations from there, when $$d>1$$.
A matrix notation answer: In matrix language rather than Einstein notation the equation $$k_ik_j\varepsilon_{ij}=2\vec k\cdot\vec u$$ could be written as $$\vec k^T\varepsilon\vec k = 2\vec k\cdot\vec u.$$ It is quite obvious out of this perspective, that there is no way to find all matrix elements of $$\varepsilon$$ from there, unless $$\varepsilon$$ is a $$1\times1$$ matrix.